Do Conformal Maps of Open Sets of the Complex Plane Always Take Boundary to Boundary? Do Conformal Maps of Open Sets of the Complex Plane Always Take Boundary to Boundary?
For instance I'm trying to create a conformal map which takes the slit open unit disk in the complex plane to the open unit disk, and takes the boundary, that is, the set $(-1,0]\cup e^{i\theta}$ for $0\leq\theta<2\pi$ to the set $e^{i\phi}$ for $0\leq\phi<2\pi$.
I currently have a conformal map which takes the slit open unit disk to the open unit disk, can I be confident that it takes boundary to boundary?
 A: Well, the relevant line in Ahlfohrs just says "For two-sided arcs the same will be true with obvious modifications. This is the second edition (1966), chapter 6, section 1.3 "Use of the Reflection principle," pages 225-226, just after the Riemann Mapping Theorem. 
So, take your domain (you need to type in something about $r e^{i \theta}$ with $0 \leq r < 1$). First, take the principal branch of the square root. That maps your domain to the semicircle $|z| < 1, \; \mbox{Re}\; z > 0.$ Then we have a domain bounded by exactly two free one-sided analytic arcs. So, Theorem 4, section 1.4, page 227, both the boundary line segment and the boundary semicircle are mapped 1-1 and analytically to the boundary circle of the unit disk. 
However, this says to me that the original slit does not have a single-valued map to the unit circle, it is evidently double-valued, just as the square root. Maybe it's just me.   
EDIT: I have it correct. See pages 18-19 in Boundary Behavior of Conformal Maps by Christian Pommerenke. What he does is switch the order, the conformal map starts in the unit disk, denoted $$f : \mathbb D \rightarrow G.$$ Note: Falcao of Atletico Madrid just scored on Real Madrid, drawn 1-1 at minute 56. OK, the boundary of $\mathbb D$ is the unit circle, denoted $\mathbb T.$

Continuity Theorem. The function $f$ has a continuous extension to $\mathbb D \cup \mathbb T$ if and only if $\partial G$ is locally
  connected.
Caratheodory Theorem. The function $f$ has a continuous and injective extension to $\mathbb D \cup \mathbb T$ if and only if
  $\partial G$ is a Jordan curve.
If $\partial G$ is locally connected but  not  a Jordan curve then
  parts of  $\partial G$ are run through several times. Some of the
  possibilities are indicated in Fig. 2.1 where points with the same
  letters correspond to each other. Note that e.g. the arcs $a_2 d_1$
  and $a_1 d_2$ are both mapped onto the segment $ad.$

